A plano-convex lens is made of refractive index 1.6. The radius of curvature of the curved surface is 60 cm. The focal length of the lens is

Options

A. 200 cm
B. 100 cm
C. 50 cm
D. 400 cm

Answer

To find the focal length of a plano-convex lens, we can use the Lensmaker's formula, which is given by:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 1: Identify the parameters
For a plano-convex lens:
- The first surface (the plane surface) has a radius of curvature \( R_1 = \infty \) (since it is flat).
- The second surface (the convex surface) has a radius of curvature \( R_2 = -60 \, \text{cm} \) (the negative sign is used because the center of curvature is on the same side as the incoming light).

### Step 2: Substitute the values into the formula
Given:
- \( \mu = 1.6 \)
- \( R_1 = \infty \)
- \( R_2 = -60 \, \text{cm} \)

Substituting these values into the Lensmaker's formula:

\[ \frac{1}{f} = (1.6 - 1) \left( \frac{1}{\infty} - \frac{1}{-60} \right) \]

### Step 3: Simplify the equation
Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:

\[ \frac{1}{f} = 0.6 \left( 0 + \frac{1}{60} \right) \]

\[ \frac{1}{f} = 0.6 \cdot \frac{1}{60} \]

\[ \frac{1}{f} = \frac{0.6}{60} \]

\[ \frac{1}{f} = \frac{1}{100} \]

### Step 4: Calculate the focal length
Taking the reciprocal gives us:

\[ f = 100 \, \text{cm} \]

Thus, the focal length of the plano-convex lens is **100 cm**.

### Final answer:
The focal length of the lens is **100 cm**.

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